Đáp án:
b) a=2M
c) \({{\text{V}}_{{H_2}}} = 4,48{\text{ lít}}\)
d) \({{\text{m}}_{dd{\text{ HCl}}}} = 36,5{\text{ gam}}\)
Giải thích các bước giải:
Phản ứng xảy ra:
\(Zn + 2HCl\xrightarrow{{}}ZnC{l_2} + {H_2}\)
Ta có:
\({n_{Zn}} = \frac{{13}}{{65}} = 0,2{\text{ mol}} \to {{\text{n}}_{HCl}} = 2{n_{Zn}} = 0,4{\text{ mol}} \to {\text{a = }}\frac{{0,4}}{{0,2}} = 2M\)
Ta có:
\({n_{{H_2}}} = {n_{Zn}} = 0,2{\text{ mol}} \to {{\text{V}}_{{H_2}}} = 0,2.22,4 = 4,48{\text{ lít}}\)
\({m_{HCl}} = 0,4.36,5 = 14,6{\text{ gam}} \to {{\text{m}}_{dd{\text{ HCl}}}} = \frac{{14,6}}{{40\% }} = 36,5{\text{ gam}}\)